The order and degree of the differential equa | vedclass

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Question

(C) Given equation: $\frac{dy}{dx} = \left(\frac{d^2y}{dx^2} + 2\right)^{1/2} + \frac{d^2y}{dx^2} + 5$ Rearranging the terms to isolate the radical: $

Vedclass · Accepted Answer

$2, 2$

Vedclass · Answer

(C) Given equation: $\frac{dy}{dx} = \left(\frac{d^2y}{dx^2} + 2\right)^{1/2} + \frac{d^2y}{dx^2} + 5$ Rearranging the terms to isolate the radical: $\frac{dy}{dx} - \frac{d^2y}{dx^2} - 5 = \left(\frac{d^2y}{dx^2} + 2\right)^{1/2}$ Squaring both sides: $\left(\frac{dy}{dx} - \frac{d^2y}{dx^2} - 5\right)^2 = \frac{d^2y}{dx^2} + 2$ Expanding the left side: $\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^2 + 25 - 2\frac{dy}{dx}\frac{d^2y}{dx^2} - 10\frac{dy}{dx} + 10\frac{d^2y}{dx^2} = \frac{d^2y}{dx^2} + 2$ Simplifying the equation: $\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 - 2\frac{dy}{dx}\frac{d^2y}{dx^2} - 10\frac{dy}{dx} + 9\frac{d^2y}{dx^2} + 23 = 0$ The highest order derivative present is $\frac{d^2y}{dx^2}$, so the order is $2$. The highest power of the highest order derivative is $2$, so the degree is $2$.

The order and degree of the differential equation $\frac{dy}{dx} = \left(\frac{d^2y}{dx^2} + 2\right)^{1/2} + \frac{d^2y}{dx^2} + 5$ are respectively

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